MSJ2e_Ch16_ISM2_June26_p708

# MSJ2e_Ch16_ISM2_June26_p708 - Chapter 16 Acids and Bases...

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Chapter 16: Acids and Bases As the reactants decompose, the concentrations of the products increase stoichiometrically, until they reach equilibrium concentrations. HA(aq) A (aq) H 3 O + (aq) conc. init. (M) 0.040 0 0 change conc. (M) – x + x + x eq. conc. (M) 0.040 – x x x At equilibrium K a = (x)(x) (0.040 - x) = 4.0 × 10 –9 Assume x is very small and: 0.040 – x 2245 0.040. 4.0 × 10 –9 = x 2 (0.040) x 2 = (4.0 × 10 –9 )(0.040) x = 1.3 × 10 –5 M = [H 3 O + ] = [A ] [HA] = 0.040 M – x = 0.040 M – 1.3 × 10 –5 M = 0.040 M (as assumed) 57. Use the method described in in the answer to Question 14.45. The equation for the equilibrium and the equilibrium expression are: C 6 H 5 COOH(aq) + H 2 O( λ ) C 6 H 5 COO (aq) + H 3 O + (aq) K a = [C 6 H 5 COO - ][H 3 O + ] [C 6 H 5 COOH] As the reactants decompose, the concentrations of the products increase stoichiometrically, until they reach equilibrium concentrations. C 6 H 5 COOH(aq) C 6 H 5 COO (aq) H 3 O + (aq) conc. init. (M) 0.050 0 0 change conc. (M) – x + x + x eq. conc. (M) 0.050 – x x x At equilibrium K a = (x)(x) (0.050 - x) = 6.3 × 10 –5 Assume x is very small and: 0.050 – x 2245 0.050. 6.3 × 10 –5 = x 2 (0.050) x 2 = (6.3 × 10 –5 )(0.050) x = 1.8 × 10 –3 M = [H 3 O + ] = [CH 3 COO ] [CH 3 COOH] = 0.20 M – x = 0.20 M – 1.9 × 10 –3 M = 0.20 M (as assumed) 58 . The equation for the equilibrium and the equilibrium expression are: CH 3 CH 2 COOH(aq) + H 2 O( λ ) CH 3 CH 2 COO (aq) + H 3 O + (aq) K a = [CH 3 CH 2 COO - ][H 3 O + ] [CH 3 CH 2 COOH] 172

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Chapter 16: Acids and Bases At equilibrium, pH = 2.93, so [H 3 O + ] = 10 –pH = 10 –2.93 = 1.2 × 10 –3 M. Since the H 3 O + ions are produced from the decomposition of C 6 H 5 COOH, an equal quantity of C 6 H 5 COO is formed, and the concentrations change in the following way: C 6 H 5 COOH(aq) C 6 H 5 COO (aq) H 3 O + (aq) conc. init. (M) 0.15 0 0 change conc. (M) – 1.2 × 10 –3 + 1.2 × 10 –3 + 1.2 × 10 –3 eq. conc. (M) 0.015 – 1.2 × 10 –3 1.2 × 10 –3 1.2 × 10 –3 K a = (1.2 × 10 - 3 )(1.2 × 10 - 3 ) (0.10 - 1.2 × 10 - 3 ) = 1.4 × 10 –5 59. Combine the methods described in Questions 31 and 51 with those described in Section 16.6 and in the answer to Question 14.45. The equation for the equilibrium and the equilibrium expression are: CH 3 NH 2 (aq) + H 2 O( λ ) CH 3 NH 3 + (aq) + OH (aq) K b = [CH 3 NH 3 + ][OH - ] [CH 3 NH 2 ] As the reactants decompose, the concentrations of the products increase stoichiometrically, until they reach equilibrium concentrations. CH 3 NH 2 (aq) CH 3 NH 3 + (aq) OH (aq) conc. init. (M) 0.23 0 0 change conc. (M) – x + x + x eq. conc. (M) 0.23 – x x x At equilibrium K b = (x)(x) (0.23 - x) = 5.0 × 10 –4 x 2 = (5.0 × 10 –4 )(0.23 – x) x 2 + 5.0 × 10 –4 x – 1.2 × 10 –4 = 0 Use the quadratic equation: (see answer to Question 14.49) x = 1.0 × 10 –2 M = [ OH ] pOH = – log[OH ] = – log(1.0 × 10 –2 ) = 2.00
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## This note was uploaded on 09/10/2008 for the course CHE 131 taught by Professor Kerber during the Spring '08 term at SUNY Stony Brook.

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MSJ2e_Ch16_ISM2_June26_p708 - Chapter 16 Acids and Bases...

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